How can we see that for an irreducible flat connection on the adjoint bundle with group $G$ on a surface $\Sigma$ , $H_2(\Sigma, ad(E))=0$? If we consider the cell decomposition of $\Sigma$ with one 2-cell $w$, and $2g$ 1-cells $a_1,b_1,...,a_g,b_g$, and one 0-cell $p$, it seems that the boundary of a parallel section over $w$ is $\partial w=a_1+b_1-a_1-b_1+...-a_g-b_g=0$, so isn't it true that $w$ is in $H_2$? Any help would be appreciated.
Source: https://link.springer.com/article/10.1007/BF02100009, page 186.
Let $\Sigma$ is a compact oriented surface (without boundary), of genus $\ge 1$. Let $\rho: \pi=\pi_1(\Sigma)\to G$ be a representation to a semisimple Lie group $G$. (Actually, all what one needs is for the Lie algebra of $G$ to admit an $Ad$-invariant nondegenerate bilinear form. In the case of semisimple Lie groups, one uses the Killing form on the Lie algebra.) Let $V$ denote the Lie algebra of $G$, regarded as an ${\mathbb R}G$-module via the adjoint representation. Let $E\to \Sigma$ be the associated vector bundle (with the fibers isomorphic to $V$) equipped with the flat connection coming from the representation. By abusing the terminology, I will use the notation $E$ also for the sheaf of horizontal sections of $E$. Then, since $\Sigma$ is aspherical, we have a natural isomorphism $$ H^\bullet(\pi, V)\cong H^\bullet(\Sigma, E). $$ Since $\Sigma$ is orientable and compact, Poincare duality (with sheaf coefficients) yields an isomorphism $$ H^0(\Sigma, E^*)\cong H_2(\Sigma, E)^* $$ where $E^*$ is the bundle associated with the coadjoint representation, the dual $(V^*, Ad^*)$ of the adjoint representation $(V, Ad)$. In view of the existence of an invariant bilinear form, we have an isomorphism $E\cong E^*$ as sheafs of horizontal sections of the respective flat bundles. Thus, in order to prove vanishing of $H_2(\Sigma, E)$ it suffices to prove vanishing of $H^0(\pi, V)$. The latter is isomorphic to the space $V^\pi$ of vectors in $V$ fixed by the representation $Ad\circ \rho$. (This is a completely general fact about group cohomology which has nothing to do with surfaces or adjoint representations. You just use the definition of group cohomology that you can find for instance in Brown's book.) By exponentiation, this space has the same dimension as the centralizer of $\rho(\pi)$ in $G$ (the set of fixed vectors is just the Lie algebra of the centralizer). If this centralizer is discrete, then we obtain $V^\pi=0$, hence, $H_2(\Sigma, E)=0$.
Incidentally, this computation also allows one to compute the dimension of $H^1(\Sigma, E)$ using the fact that $$ \chi(\Sigma)\cdot \dim G= \chi(\Sigma)\cdot \dim V= \chi(\Sigma, E)= \dim H^0(\Sigma, E) - \dim H^1(\Sigma, E) + \dim H^2(\Sigma, E). $$
Since $\dim H^0(\Sigma, E)=\dim H^2(\Sigma, E)=0$, we obtain $$ \dim H^1(\Sigma, E)= (2p-2)\dim G, $$ where $p$ is the genus of $\Sigma$.